Quarterly Journal of the Royal Meteorological Society Q. J. R. Meteorol. Soc. 00: 1–12 (2011)
The forces of inertial oscillations
Jeffrey J. Early
Oregon State University, Corvallis, Oregon, USA∗Correspondence to: Jeffrey J. Early, NorthWest Research Associates, 4118 148th Ave NE, Redmond, WA 98052.
E-mail: [email protected]
By starting with a free particle and successively adding
constraints, it is shown that the free motion of a particle
constrained to the earth’s surface is inertial, despite statements
in the literature, and that an observer on this particle would
not measure a force tangent to the Earth’s surface. However, if
the observer extended his measurements to include the direction
normal to the Earth’s surface, he would detect an oscillating force.
Copyright c� 2011 Royal Meteorological Society
Key Words: inertial oscillations
Received . . .
Citation: . . .
1. Introduction
The primitive equations in oceanography and atmospheric
science describe a rich set of physical phenomena, but
their complexity makes analytical solutions exceptionally
rare. Only by simplifying the equations do we hope
to find such solutions and fully explain the forcing
involved in the resulting solution. The inertial oscillation
problem exemplifies this approach by reducing the primitive
equations to describe the motion of a particle confined
to the surface of the earth. While inertial motions are
a ubiquitous facet of ocean currents (Gill 1982; Pollard
and Millard 1970), the interest in this problem lies also
in its simplification of the underlying forcing. Indeed,
when mathematician and meteorologist Francis John Welsh
Whipple (Whipple 1917) stated the problem in 1917 he
wrote “...the author considers that a knowledge of the
motion of such a particle will prove a useful preliminary
to a proper understanding of the more complicated motion
which actually occurs in winds, where the air particles
have other forces besides that of gravity acting upon them.”
Nearly a century later the exact solutions for the particle
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2 J. J. Early
constrained to the earth have been found (Pennell and
Seitter 1990; Paldor and Sigalov 2001) but do not appear
to be well known and the forces responsible for the motion
have continued to be debated and discussed (Stommel
and Moore 1989; Durran 1993; Ripa 1997; Persson 1998;
Phillips 2000).
This manuscript corrects a fundamental and widespread
misconception concerning inertial oscillations and whether
or not inertial oscillations are accurately described by
the term ‘inertial’. It’s important to note that this is not
merely an issue of semantics. The name ‘inertial oscillation’
suggests that the observed oscillatory motion of the particle
is explained entirely by inertial forces, apparent only
because of our choice of reference frame. We can reframe
this issue as a simple thought experiment by asking whether
or not an observer inside a small laboratory set in a
perfect ‘inertial oscillation’ could design an experiment to
distinguish his laboratory from an inertial frame. In a more
practical sense, we are asking whether an accelerometer
set in perfect inertial motion measure accelerations? If the
accelerometer does not measure an acceleration then the
name ‘inertial oscillation’ is appropriate, but by contrast if
an acceleration is detected, then the name is fundamentally
wrong and should be changed.
There is a seductively simple, but incorrect, proof which
appears to show that there is in fact a force in the
inertial frame. The f -plane approximation to the horizontal
equations of motion are simply
dvr
dt+ 2ω × vr = 0 (1)
where vr is the two-dimensional velocity vector of the
according to the observer on the earth and ω is the angular
frequency of rotation of the earth. By simple kinematic
computation the relationship between acceleration in the
rotating and fixed frame is given by
dvf
dt=
dvr
dt+ 2ω × vr + ω × (ω × r). (2)
If we take equation (2) and apply equation (1) we see that
in the fixed frame,
dvf
dt= ω × (ω × r). (3)
Because vf is the velocity in the fixed frame, then by
Newton’s second law equation (3) shows there to be a force
acting on the particle. Although the force in equation (3)
appears to be centrifugal, Durran (1993) correctly points
out that the term is actually a component of gravity. From
this we might conclude that because there is a force in the
fixed frame, the particle’s motion is not inertial. This result
suggests a profound difference from the original claim
that an accelerometer would not detect any accelerations.
By this reasoning it is now repeated in the literature and
the community that these oscillations are not inertial, in
contradiction with the claim that an accelerometer would
not detect any accelerations.
Starting with free particle and successively adding
constraints we will show that the particle’s motion is in
fact inertial and the accelerometer trapped in inertial motion
would not measure an acceleration. The ‘force’ identified
in equation (3) is a component of gravity, fundamentally
indistinguishable from other inertial forces that result from
our choice of reference frame. General relativity tells us
that an astronaut in orbit around the earth can measure
no accelerations, despite the curvature of his orbit and the
presence of gravity. For the same reason a particle in an
inertial oscillation on the surface of the earth will measure
an acceleration only in the vertical direction, but not in the
direction of its motion, due to changes in the strength of
the normal force throughout the course of an oscillation.
Inertial oscillations are truly inertial in the local horizontal
and therefore deserving of the name. On the other hand, it
will also be shown that the name inertial oscillation suffers
from ambiguity and that a more descriptive name can be
devised based on the problem constraints.
To understand the forcing we will use the Lagrangian
(Lagrange function) to help write down and understand
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The forces of inertial oscillations 3
the forces involved. The advantage to this technique is
its simplicity: small changes in our assumptions about
the system effortlessly show changes in the forcing.
Starting with the description of a free particle in spherical
coordinates and adding a new assumption at each step, we
examine two other systems nearly identical to the particle
on a earth: a particle in orbit and a particle constrained to
a sphere. The particle in orbit introduces gravity into the
problem without the complication of constraints and then
by constraining the particle to the sphere, the effects of the
constraint force are more easily isolated. By considering
these two other systems first, we will finally highlight the
subtle differences that arise in physical systems on the earth
and have a clear understanding of source of each force. For
completeness, the exact solutions – extending the work of
Pennell and Seitter (1990) and Paldor and Sigalov (2001) –
are presented in the Appendix.
2. Free Particle, Inertial Frame
The Lagrangian for a conservative mechanical system is the
kinetic energy minus the potential energy, L = KE − PE.
We find the path that minimizes the action, S =�L dt,
using the Euler-Lagrange equations, ddt
�∂L∂q
�= ∂L
∂q , and
recover the equations of motion exactly as if we’d written
out the forces explicitly using Newton’s second law.
We use the coordinate system (φI , θ, r) in the inertial
frame where φI points eastward in the direction of
increasing longitude and θ points northward, in the direction
of increasing latitude. To an observer standing on the Earth,
these coordinates locally look like (x, y, z): east, north and
up. The Lagrangian for a free particle is the kinetic energy
of the particle,
LIfree =
1
2r2 +
1
2r2θ2 +
1
2r2φI
2cos2 θ (4)
where dot is used to indicate the time derivative, e.g., r =
drdt . By applying the Euler-Lagrange equations to (4), the
equations of motion are found to be
aφI
aθ
ar
≡
rφI cos θ + 2rφI cos θ − 2rφI θ
rθ + 2rθ + rφ2I sin θ cos θ
r − rθ2 − rφ2I cos
2 θ
=
0
0
0
,
(5)
which we’ve also used to define the acceleration vector
a, in terms of the coordinates (φI , θ, r) and their time
derivatives. Equation (5) is written to resemble Newton’s
second law a = f where a is the acceleration and f is the
specific force (zero, in this case). Although these are the
equations of motion for a free particle, they appear rather
complicated because they’re written in a coordinate system
not well suited for the problem, spherical coordinates. It
is important to keep in mind that all of these terms are
necessary simply to describe an unforced, and therefore
unaccelerated, particle moving in a straight line. Only
additional constraints added to the Lagrangian will result
in terms that appear as forces.
3. Free Particle, Rotating Frame
In order to identify the so-called “inertial forces”, we need
to transform equation (4) into the rotating frame. We apply
the transformation φI �→ φ+ ωt where ω is the angular
rotation frequency of the earth and φ is the new longitudinal
coordinate. The Lagrangian becomes,
LRfree =
1
2r2 +
1
2r2θ2 +
1
2r2
�φ2 + 2ωφ+ ω2
�cos2 θ,
(6)
from which we obtain the following equations of motion,
aφ
aθ
ar
=
−2rω cos θ + 2rθω sin θ
−2rφω sin θ cos θ − rω2 sin θ cos θ
2rφω cos2 θ + rω2 cos2 θ
. (7)
To the rotating observer believing he is in an inertial frame,
everything on the left-hand-side of the equations appear
as appropriate acceleration terms for his chosen spherical
coordinate system, while everything on the right-hand-side
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4 J. J. Early
appear to be forces. These are the so-called “inertial forces”.
The four terms on the right containing a velocity are the
Coriolis force, while the two terms only dependent on
position are the centrifugal force.
Imagine that vector rI describes the motion of a
spaceship far between the stars, which is therefore
experiencing no forcing, and satisfies equation (5). The
rotating observer, in contrast, describes the motion of the
spaceship with vector rR satisfying equations (7). In a
practical sense what makes these forces in equation (7)
inertial is that when a person on the spaceship measures
his acceleration with an accelerometer, he will find his
acceleration to be zero. By Newton’s second law the person
in the spaceship would correctly conclude no forces are
acting on him, despite the rotating observer’s belief to the
contrary.
4. Central Gravitational Field, Rotating Frame
Let’s extend the free particle Lagrangian of equation (6) by
including a central gravity field similar to the earth’s,
LRgravity =
1
2r2 +
1
2r2θ2
+1
2r2
�φ2 + 2ωφ+ ω2
�cos2 θ +
GM
r.
(8)
The equations of motion for this system differ from equation
(7) only by the addition of a gravitational force included in
the radial acceleration, ar,
aφ
aθ
ar
=
−2rω cos θ + 2rθω sin θ
−2rφω sin θ cos θ − rω2 sin θ cos θ
2rφω cos2 θ + rω2 cos2 θ − GM
r2
. (9)
The solutions to (9) are also well known and include
motions like the International Space Station’s (ISS) orbit
path and geosynchronous satellites, figure (1).
This system too only involves “inertial forces.” From
Einstein’s equivalence principle we know that we could not
conduct an experiment to distinguish between an elevator
at rest on the earth and an elevator accelerating by rocket
at 9.8 m s−2. So while Newton would have argued that
we introduced a force of gravity into equation (9), by
the equivalence principle we know that this is really
just another “inertial” term. As a practical consequence,
astronauts aboard the ISS do not detect acceleration with
their accelerometers. Because they are in free fall, they
are in an inertial reference frame, locally indistinguishable
from the reference frame of the inertial observer of section
(3) without a central gravitational field. This reflects an
evolution in our understanding of forces from general
relativity because we cannot design a local experiment to
distinguish between the cases with and without this term.
Gravity is now more appropriately considered a description
of the local geometry of the problem, and not a force. This
is an important point at the heart of general relativity, see
chapter 4.3 of Wald (1984) or chapter 1 of Misner et al.
(1973).
By this reasoning then, the oscillations of satellites
around the earth or oscillations of satellites in near
geosynchronous orbit could justly be named “inertial
oscillations.”
5. Particle on a Rotating Sphere
We can constrain the particle to the surface of a sphere
simply by setting r = R. However, a more fruitful
technique is to apply our constraint with a Lagrange
multiplier, denoted λRsphere, so that we analyze the resulting
force required to keep the particle on the surface of the
sphere. λRsphere is treated as variable just like φ, θ and r, but it
multiplies the constraint, namely that r −R = 0. With this
addition, the Lagrangian becomes
LRsphere =
1
2r2 +
1
2r2θ2 +
1
2r2
�φ2 + 2ωφ+ ω2
�cos2 θ
+GM
r+ λR
sphere(r −R).
(10)
In the inertial frame equation (10) would describe geodesic
motion on the surface of a sphere, the path of the great
circles. This system was analyzed by McIntyre (2000) in
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The forces of inertial oscillations 5
Figure 1. Three example particle paths in a central gravitational field as viewed from a rotating reference frame. Panel a) shows an orbit that approximatesthe International Space Station, b) is a small deviation from geosynchronous orbit and c) is a geosynchronous orbit with an inclination. All three areinertial and exhibit oscillatory motion.
both the rotating and the inertial frame. The equations of
motion now include a fourth equation using λRsphere as a
variable in the Euler-Lagrange equations,
aφ
aθ
ar
r
=
−2rω cos θ + 2rθω sin θ
−2rφω sin θ cos θ − rω2 sin θ cos θ
2rφω cos2 θ + rω2 cos2 θ − GMr2 + λR
sphere
R
.
(11)
By applying the constraint condition r = R and using
the definitions in equation (5), we can now use the ar
component of equation (11) to find the force of constraint,
λRsphere =
gravity� �� �GM
R2−
coriolis� �� �2Rφω cos2 θ−
centrifugal� �� �Rω2 cos2 θ
−Rθ2 −Rφ2 cos2 θ� �� �geometric
.
(12)
The force of constraint in equation (12) is synonymous
with the term normal force (represented FN) because it
is necessarily perpendicular to the resulting motion. The
normal force is a real force and stands in as a proxy
for the collection of electromagnetic forces that keep our
particle from plunging into the depths of this spherical earth.
The accelerations caused by this force are detectable by
our accelerometer. The first term is exactly opposite and
equal to the force of gravity, completely negating its effect.
The next two terms in equation (12) oppose components
Fg
FN
Fcentrifugal
!
FNet
Figure 2. Force diagram for a particle initially at rest on a rotating sphere.The net force points towards the equator, but only the normal force, FN,is a real, measurable force.
of the inertial forces, the Coriolis and centrifugal force.
This real force that opposes the inertial centrifugal force
is called the centripetal force. The last two terms are
perhaps unexpected. They represent the constraint force
applied against the particle’s inertia preventing it from
moving in a straight, unaccelerated path. These two terms
would be present even in the inertial frame (ω = 0) and
without gravity (G = 0); they distinguish geodesic motion
in Euclidean space from geodesic motion on a sphere and
could be called geometric forces.
Consider the forces acting on the particle initially at rest
by setting (φ, θ, r) = (0, 0, 0) in equation (11). In this case
the only forces acting on the particle are gravity, the normal
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6 J. J. Early
force, and the centrifugal force, see figure (2). However,
we can see from equation (11) that a component of the
centrifugal force still points in the −θ direction, pushing the
particle towards the equator. This net force is exactly why
we know that this Lagrangian does not correctly describe a
particle on the earth’s surface (objects initially at rest on the
earth’s surface do not roll equatorward).
The resulting motion shown in figure (3) is again
periodic, so could this too be described as inertial
oscillations? We have established that there exists a force
of constraint, but how would the observer moving with the
particle experience this motion? Assume the observer starts
initially at rest with respect to the rotating surface. We can
see from figure (3) that the observer will oscillate about the
equator, drifting slowly westward. Because there are only
inertial forces in the φ and θ directions, the observer will
not detect any accelerations in those directions. However,
the observer’s local vertical acceleration will show change,
directly computable with equation (12). There will be a
constant acceleration detected opposite and equal to gravity,
but in addition there is a position dependent centripetal force
as well as the other velocity dependent geometric forces
in equation (12). To the three-dimensional observer, any
motion described by this system has detectable forces and
therefore is not inertial motion. However, the spirit of the
problem is to restrict the motion, and perhaps therefore also
the observations, to only two dimensions. In the remaining
two dimensions the motion only contains inertial forces and
it seems reasonable to describe the oscillatory motion as
inertial oscillations.
6. Particle on Earth
In this final case we want to consider a particle under
the influence of gravity and constrained to the surface of
the earth. The gravitational potential is parameterized as
the more general G(r, θ) instead of the potential for a
point mass (and sphere), −GMr . The surface of the earth
is described by some function f(r, θ) and therefore our
constraint is described by f(r, θ) = k. The Lagrangian in
the rotating frame is
LRearth =
1
2r2 +
1
2r2θ2 +
1
2r2
�φ2 + 2ωφ+ ω2
�cos2 θ
−G(r, θ) + λRearth (f(r, θ)− k) .
(13)
This looks identical to equation (10) except for the
generalizations of the gravitational potential and the
constraint force. The key difference, shown in Ripa (1997),
is that
Vgeo(r, θ) ≡ G(r, θ)− 1
2r2ω2 cos2 θ (14)
must be equal to a constant at the surface of the earth.
According to NIMA (2000) Vgeo is an effective potential
called the total gravity potential and a surface of constant
potential is a geopotential surface, or geop. In NIMA (2000)
it also stated that “the geoid is that particular geop that is
closely associated with the mean ocean surface.” So this
means that Vgeo describes the surface of the earth, which
is exactly what we said f(r, θ) does! This should simplify
things.
The idea that the earth’s surface is defined by the
geopotential can be understood by considering the previous
example of the particle on the rotating sphere. If the earth
were a sphere as described by the system (10), then all
particles on its surface would roll towards the equator. So
if we imagine the earth were entirely fluid and in static
equilibrium everything would have already rolled towards
the equator; its shape would have relaxed to this minimal
potential energy state. The reality is that over the oceans it’s
pretty close.
However, now we have a problem with our coordinates. A
good set of coordinates for this problem would be defined
with one basis vector normal to the surface and the other
two basis vectors tangent to the surface. Our spherical
coordinates do this for a sphere, but not for a geopotential
surface which is clearly a function of both θ and r. The
vector φ points tangent to the surface, but θ and r have
components both tangent and normal to the surface. If we
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The forces of inertial oscillations 7
Figure 3. Three example particle paths on a sphere as viewed from a rotating reference frame of an eastward rotate sphere. Panel a) depicts the path ofa particle released from rest in the rotating reference frame, which subsequently drifts to the west. Panels b) and c) show different choices of initiallyeastward velocities. All three are inertial tangent to the surface, and oscillatory.
approximate our geopotential as an ellipse, and then use
elliptic coordinates, we can achieve the desired results.
This turns out not to change the form of the Lagrangian
significantly. The key observation is that if we use a good
set of coordinates where say, ξ is normal to the geopotential,
then Vgeo is entirely a function of ξ. The change in direction
of the forcing from the difference between the geometry
of a spheroid and an ellipsoid is negligible (Gill 1982). At
this point we will then redefine our coordinates so that r is
normal to the geopotential and θ is perpendicular to both
r and φ. This also allows us to restate our constraint as
constant r, rather than constant Vgeo. The rotating observer
standing on the earth experiences −∂Vgeo∂r as the total force of
gravity, so we use equation (14) and neglecting those small
geometric deviations to rewrite equation (13) as
LRearth =
1
2r2 +
1
2r2θ2 +
1
2r2
�φ2 + 2ωφ
�cos2 θ
− Vgeo(r) + λRearth (r −R) ,
(15)
where R is the approximate radius of the earth.
At this point if we were to ignore the force of constraint
and simply proceed to set r = R we would recover the same
Lagrangian as in Ripa (1997), and therefore produce the
same force diagrams drawn in figure 1 of both Ripa (1997)
and Durran (1993). However, let’s proceed with the force
of constraint still under consideration and apply the Euler-
Lagrange equations to examine the resulting force. In the
rotating frame we find
aφ
aθ
ar
r
=
−2rω cos θ + 2rθω sin θ
−2rφω sin θ cos θ
2rφω cos2 θ − ∂Vgeo∂r + λR
earth
R
. (16)
Equation (16) looks almost identical to the equations of
motion on a sphere, equation (11). The centrifugal and
gravitational force in the radial component of equation
(11) are combined in the definition of Vgeo. However, the
centrifugal force in the θ component of equation (11) is
missing from equation (16). Qualitatively at least this looks
a lot more like the surface of the earth as we know it:
objects initially at rest do not roll towards the equator. The
constraint force is
λRearth =
∂Vgeo
∂r
�����r=R
− 2Rφω cos2 θ −Rθ2 −Rφ2 cos2 θ,
(17)
which appears nearly identical to the constraint force on the
sphere, equation (12). For the particle initially at rest, the
constraint force evaluates to
λRearth =
∂Vgeo
∂r
�����r=R
=∂G
∂r
�����r=R
−Rω2 cos2 θ (18)
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8 J. J. Early
Fg
FN
Fcentrifugal
Figure 4. Force diagram for a particle initially at rest on the earth. Allforces balance, so the particle will remain at rest in this reference frame.The outline of a sphere is shown in gray.
and is depicted in figure (4). Because the constraint force
is the only real force in this system, the motion is inertial in
the φ and θ directions, just as with the particle on the sphere.
So where’s the discrepancy with the derivation in section 1?
Consider the Lagrangian in the inertial frame, but without
using equation (14) to replace the gravitational potential
with the total gravity potential,
LIearth =
1
2r2 +
1
2r2θ2 +
1
2r2φI
2cos2 θ
−G(r, θ) + λIearth (r −R) .
(19)
Applying the Euler-Lagrange equations we find that
aφI
aθ
ar
r
=
0
− 1r∂G∂θ
−∂G∂r + λI
earth
R
. (20)
Equation (20) shows that unlike the case of the sphere,
gravity has a component tangent to the surface and in
fact, from equation (14), we know the magnitude of this
component is exactly rω2 sin θ cos θ. The gravitational
forces of equation (20) are however, not real forces and it
is only the constraint force λIearth found in the r direction
that is measured by the accelerometer. Solving for λIearth in
equation (20),
λIearth =
∂G
∂r−Rθ2 −RφI
2cos2 θ, (21)
we see that the constraint force balances only the radial
component of gravity, r, and not its horizontal components.
A particle at rest in the rotating frame, (φI , θ, r) = (ω, 0, 0),
has a normal force of
λRearth =
∂G
∂r
�����r=R
−Rω2 cos2 θ. (22)
Equations (18) and (22) are identical, meaning the rotating
observer and inertial observer predict the same reading
on the accelerometer. The forces as they appear from the
inertial frame are drawn in figure (4). For the inertial
observer having a non-zero net force looks correct. The
particle is rotating around the polar axis at speed ω and
therefore is accelerating inward.
Following the incorrect logic of the proof in section
1, consider what would happen if we believed that the θ
component of gravity in equation (20) really was a force.
This would suggest that accelerometer measurement by
an observer standing on the earth would point not in the
local vertical, but would point slightly equatorward! This
is certainly not our experience and so we must reject the
notion that gravity is a true force.
The accelerometer trapped in inertial oscillations would
therefore not detect any motion in the φ or θ directions.
However, like the particle on the sphere, the accelerometer
would detect changes in acceleration in the local vertical
due to the changing magnitude of the normal force. So
again, we could justly call these oscillations are inertial
if we restrict measurements to the two dimensions of the
earth’s surface. Representative solutions are shown in figure
(5)
Notice that the constraint force (17) provides enough
information that an observer in inertial motion measuring
acceleration in the vertical would be able to deduce his
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The forces of inertial oscillations 9
motion. The first term in the constraint force is constant,
opposite and equal to the total gravity, and the last two
terms are proportional to the energy, and are therefore also
constant. The frequency and magnitude of the oscillations
of the second term would allow the observer to determine
the latitude and speed of his motion at any instant, given the
solutions shown in the appendix.
7. Conclusions
The name “inertial oscillations” turns out not to be
so bad after all. The “force” shown to exist in the θ
direction is gravity and by Einstein’s equivalence principle
it is therefore not measurable by an accelerometer. This
additional term should not strictly be considered a force,
but in fact a description of the local geometry. In our
daily lives we often mistakenly think of gravity as the
force we feel against our feet, but the reality is that that
it’s the earth pushing us up. The result of this is that an
accelerometer trapped in an inertial oscillation would not
detect any accelerations in the direction tangent to earth’s
surface and only measurement in the local vertical would
indicate that the particle is being forced. So given the
lack of observable forces the oscillations described by the
Lagrangian in equation (15) are inertial.
Is inertial oscillation really an appropriate name for
this phenomenon? In all four systems we examined, the
motion could be described as inertial and all four systems
exhibit oscillations, so the name “inertial oscillation” isn’t
particularly descriptive. Durran (1993) suggested “constant
angular momentum oscillation”, but this too suffers from
ambiguity. Because φ does not explicitly enter into any of
the four Lagrangians we consider, this means that angular
momentum is conserved in all four cases. The key feature
distinguishing oscillations on the earth from the other three
systems is the constraint of the particle to the geopotential.
In order to satisfy equation (14) the direction of the total
gravity force and the gravitational force always differ by a
centrifugal component, resulting in an apparent net force
in the inertial frame. This will always be a feature of a
geopotential of the Earth or any other rotating planet and
therefore the name “geoinertial oscillation” is perhaps more
descriptive.
Acknowledgement
Thanks to Roger M. Samelson and Jonathan Lilly
for their insightful comments and suggestions. Much
thanks to Emily Shroyer for her insight in choosing
an appropriately descriptive name. This research was
supported by the National Science Foundation, Award
0621134 and 1031286.
A. Exact Solutions
Numerical solutions to the inertial oscillation problem
have been previously considered (Paldor and Killworth
1988), but the exact solutions were first found by directly
integrating the equations of motion (Pennell and Seitter
1990) and then later by use of action-angle variables (Paldor
and Sigalov 2001). The exact solutions can also be found
by directly integrating the conserved quantities implied by
the Lagrangian in equation (15), which is the approach
taken here. From the three different solutions for the path
(φ(t), θ(t)), we can also compute, for the first time, the
particle velocity, (u(t), v(t)) for the different solutions in
closed form. A Matlab script to compute the exact solutions
for particle path, velocity, and frequency of oscillation for
all three cases is available at http://jeffreyearly.com.
Starting with equation (15), but applying the constraint
that r = R, the Lagrangian simplifies to
LRearth =
1
2R2θ2 +
1
2R2
�φ2 + 2ωφ
�cos2 θ. (23)
Equation (23) has two conserved quantities, energy E ≡
q ∂L∂q − L, and angular momentum L ≡ ∂L
∂φ. The energy can
be written as
E =1
2R2θ2 +
1
2R2 cos2 θφ2,
Copyright c� 2011 Royal Meteorological Society Q. J. R. Meteorol. Soc. 00: 1–12 (2011)
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10 J. J. Early
Figure 5. Three example particle paths on the earth as viewed from a rotating reference frame. Just as with the sphere, all three are inertial tangent tothe surface and exhibit oscillatory motion.
or in the more familiar E =�u2 + v2
�/2 if we define
velocities (u, v) =�Rφ cos θ, Rθ
�. The angular momen-
tum of the particle
L = R2 cos2 θ(φ+ ω) (24)
can be used to eliminate φ from the energy equation,
E =1
2
L2
R2 cos2 θ+
1
2R2θ2 +
1
2R2ω2 cos2 θ − ωL. (25)
Equation (25) is a first order ordinary differential equation
that can be used to compute θ(t) by direct integration. After
finding θ(t), φ(t) can be found by integrating equation (24).
Solving for θ in equation (25) and then integrating,
ωt =
� θ
θ0
cos θ�2(E+ωL)
ω2R2 cos2 θ − L2
ω2R4 − cos4 θdθ. (26)
Choosing initial conditions such that the particle is
launched with an initially eastward trajectory, (u0, v0) =
(Rφ0 cos θ0, 0), allows us to factor the quartic polynomial
under the radical. Writing the conserved quantities E and
L in equation (26) in terms of these initial conditions, the
integral becomes
ωt =
� θ
θ0
cos θ�(cos2 θ − cos2 θ0) ((1 + uφ)2 cos2 θ0 − cos2 θ)
dθ
(27)
where we’ve defined uφ = φ0/ω, the ratio of the initial
velocity of the particle to the tangential velocity of the earth.
The substitution s = sin θ/ sin θ0 reduces equation (27)
to a standard form for inverse elliptic integrals,
sin θ0ωt =
� sin θsin θ0
1
ds�(1− s2)
�s2 − k�2
� (28)
where k�2 = (1− (1 + uφ)2 cos2 θ0)/ sin2 θ0. The par-
ameter k� is called the complementary modulus and is
related to the modulus by k2 + k�2 = 1 (Byrd et al. 1971).
The modulus is therefore k2 = cot2 θ0(2uφ + uφ2), but it
can also be written in terms of more standard oceano-
graphic constants, u0 = φ0R cos θ0, f0 = 2ω sin θ0, and
β = 2ω cos θ0/R. Using these values,
k2 = 4
�u0β
f20
+u20
R2f20
�.
The modulus separates inverse elliptic integrals from
inverse trigonometric integrals. Notice that if the modulus
were 0, equation (28) would be an inverse trigonometric
integral, whereas if k = 1, equation (28) would be the
integral for an inverse hyperbolic function. The modulus
depends on both the energy via the u20 term and the angular
momentum via the u0 term. The modulus is dependent on
θ0, the maximum latitude achieved by the particle rather
than θmid, the mid-latitude where u = 0, but using the two
conservation properties one can express k in terms of θmid if
desired.
Copyright c� 2011 Royal Meteorological Society Q. J. R. Meteorol. Soc. 00: 1–12 (2011)
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The forces of inertial oscillations 11
A.1. Case k2 < 1
When k2 < 1, the integral in equation (28) is exactly the
definition of the inverse elliptic function dn−1(u, k) (Byrd
et al. 1971) and therefore
θ1(t) = sin−1 (sin θ0 dn(ω�t, k)) , (29)
where ω� = sin θ0ω. The solution to φ(t) is found using
equation (24),
φ(t) = cos2 θ0(φ0 + ω)
�1
1− sin2 θ0 dn2(ω�t)
dt− ωt.
(30)
We can write dn2 u in terms of sn2 u using the relation
dn2 u = 1− k2 sn2 u to rewrite equation (30),
φ(t) =1 + uφ
sin θ0
� ω�t
0
ds
1 + k2 tan2 θ0 sn2 s− ωt. (31)
The integral is exactly the definition of an elliptic integral
of the third kind (Byrd et al. 1971) and thus,
φ1(t) =(1 + uφ)
sin θ0Π�am (ω�t) ,−k2 tan2 θ0, k
�− ωt
(32)
where am(u, k) is the Jacobi amplitude such that,
e.g., sin(am(u, k)) = sn(u, k). The solution (θ1(t),φ1(t))
oscillates between cos−1 ((1 + uφ) cos θ0) and θ0 and has
a slow westward drift that can be computed by take the
integral of equation (30) over one period, similar to using
the action-angle variables in Paldor and Sigalov (2001).
The frequency of oscillation occurs at f = 2ω sin θ using
the f -plane approximation, but can be computed exactly
by noting that the dn function is 2K(k) periodic where
K(k) is the complete elliptic function and dependent on the
modulus. Doing this we find that,
f1(θ0) =πω sin θ0K(k)
. (33)
The u component of velocity follows from equation (24)
by noting that
R cos θ(t) (u(t) + ωR cos θ(t)) = R2 cos2 θ0�φ0 + ω
�
and then solving for u(t), while v component is found by
differentiating θ(t) with respect to t. The velocity for the
first solution is therefore
u1(t)
v1(t)
=1�
1 + k2 tan2 θ0 sn2 (ω�t, k)
·
u0 − 2
�u0 +
u20
R2β
�sn2 (ω�t, k)
−2�u0 +
u20
R2β
�sn (ω�t, k) cn (ω�t, k)
(34)
A.2. Case k2 = 1
If we set k2 = 1, then equation (28) reduces to the usual
integral for the inverse sech function, which can also be
seen from equation (29) because dn(u, 1) = sech(u). The
solution for θ(t) is therefore,
θ2(t) = sin−1 (sin θ0 sech(ω�t)) , (35)
and φ(t) follows from the same integration of equation (24),
φ2(t) = ω(1− cos θ0)t+ tan−1 (tan θ0 tanh (ω�t)) .
(36)
This is clearly a special case that requires a launch of
the particle at precisely the right velocity, u = ωR(1−
cos θ0). In this case the particle hits the equator, essentially
remaining in orbit around the equator for its lifetime.
Using the same method as the first case, the velocity of
the particle for the second solution is
u2(t)
v2(t)
=1�
1 + k2 tan2 θ0 tanh2 (ω�)
·
u0 − f20
2β tanh2 (ω�t)
− f20
2β tanh (ω�t) sech (ω�t)
.
(37)
Copyright c� 2011 Royal Meteorological Society Q. J. R. Meteorol. Soc. 00: 1–12 (2011)
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12 J. J. Early
A.3. Case k2 > 1
The modulus k of a Jacobi elliptic functions is assumed to
be 0 < k < 1, but in this case k exceeds 1. We can work
around this issue by using the reciprocal of the modulus,
k−1, instead. Because dn(u, k) = cn�ku, k−1
�, it follows
that
θ3(t) = sin−1�sin θ0 cn
�kω�t, k−1
��, (38)
and the φ(t) solution for this case proceeds in the same way
as the k2 < 1 case,
φ3(t) =(1 + uφ)
k sin θ0Π�am (kω�t) ,− tan2 θ0, k
−1�− ωt.
(39)
The solution (θ3(t),φ3(t)) oscillates between θ0 and −θ0,
and has either an eastward or an westward drift depending
on the initial conditions.
The cn function is 4K(k) periodic and therefore the
frequency of oscillation is,
f3(θ0) =kπω sin θ02K (k−1)
. (40)
Finally, the velocity of the particle is
u3(t)
v3(t)
=1�
1 + tan2 θ0 sn2 (kω�t, k−1)
·
u0 − f20
2β sn2�kω�t, k−1
�
− f20
2β k sn�kω�t, k−1
�cn
�kω�t, k−1
�
.
(41)
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